Topic 6: Circular motion and gravitation 6.1 – Circular motion

Topic 6: Circular motion and gravitation 6.1 – Circular motion
Guidance: • Banking will be considered qualitatively only Data booklet reference: • v = r • a = v 2 / r = 4 2r / T 2 • F = mv 2 / r = m2r © 2006 By Timothy K. Lund 1

Centripetal force and acceleration What force must be applied to Helen to keep her moving in a circle? How does it depend on the Helen’s radius r ? How does it depend on Helen’s velocity v? How does it depend on Helen’s mass m? r m v © 2006 By Timothy K. Lund On the next pass, however, Helen failed to clear the mountains.

Centripetal force and acceleration A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v. Observe that the velocity vector is always tangent to the circle. Note that the magnitude of the velocity vector is NOT changing. Note that the direction of the velocity vector IS changing. Thus, there is an acceleration, even though the speed is not changing! r blue v red x y © 2006 By Timothy K. Lund v r

Centripetal force and acceleration To find the direction of the acceleration (a = v / t ) we observe two nearby snapshots of the particle: The direction of the acceleration is gotten from v = v2 – v1 = v2 + (-v1): The direction of the acceleration is toward the center of the circle - you must be able to sketch this. r blue v red v1 v2 v2 x y © 2006 By Timothy K. Lund v1 v2 -v1 v v1 -v1 v FYI Centripetal means center-seeking.

Fc Centripetal force and acceleration How does centripetal acceleration ac depend on r and v ? To explore this we define the centripetal force Fc: Picture yourself as the passenger in a car that is rounding a left turn: The sharper the turn, the harder you and your door push against each other. (Small r = big Fc.) The faster the turn, the harder you and your door push against each other. (Big v = big Fc.) Fc = mac centripetal force © 2006 By Timothy K. Lund

Centripetal force and acceleration PRACTICE: For each experiment A and B, label the control, independent, and dependent variables. no change A manipulated B Fc ac v r Fc ac v r Fc ac v r Fc ac v r manipulated no change © 2006 By Timothy K. Lund responding responding  CONTROL: r  CONTROL: v  INDEPENDENT: v  INDEPENDENT: r  DEPENDENT: Fc , ac  DEPENDENT: Fc , ac

Centripetal force and acceleration We know the following things about ac: If v increases, ac increases. If r increases, ac decreases. From dimensional analysis we have What can we do to v or r to “fix” the units? This is the correct one! ac = v r first guess formula v r ? m s2 m/s m ? 1 s ac =  = = © 2006 By Timothy K. Lund ac = v 2 / r centripetal acceleration ? ? v2 r m s2 m2/s2 m m s2 ac =  = =

Solving centripetal acceleration and force problems Fc = mac centripetal force ac = v 2 / r centripetal acceleration EXAMPLE: A 730-kg Smart Car negotiates a 30. m radius turn at 25. m s-1. What is its centripetal acceleration and force? What force is causing this acceleration? SOLUTION: ac = v2 / r = 252 / 30 = 21 m s-2. Fc = mac = (730)(21) = n. The centripetal force is caused by the friction force between the tires and the pavement. © 2006 By Timothy K. Lund

Period and frequency The period T is the time for one complete revolution. The frequency f (measured in Hz or cycles / s) is defined as how many cycles (oscillations, repetitions, revolutions) occur each second. Since period T is seconds per revolution, frequency must be 1 / T. © 2006 By Timothy K. Lund f = 1 / T relation between T and f or T = 1 / f EXAMPLE: Find the period and the frequency of a day. SOLUTION: The period is T = (24 h)(3600 s h-1) = s. The frequency is f = 1 / T = 1 / = 1.1610-5 Hz.

Period and centripetal acceleration Sometimes the period of a revolution is given, rather than a velocity. One revolution is one circumference C = 2r. Therefore v = distance / time = 2r / T. Thus v 2 = 4 2 r 2 / T 2 so that ac = v 2 / r = 4 2 r 2 / T 2r = 4 2 r / T 2. © 2006 By Timothy K. Lund ac = v 2 / r centripetal acceleration ac = 4 2 r / T 2

Solving centripetal acceleration and force problems ac = v 2 / r centripetal acceleration ac = 4 2 r / T 2 EXAMPLE: Albert the 2.50-kg physics cat is being swung around by a string harness having a radius of 3.00 meters. He takes 5.00 seconds to complete one fun revolution. What are ac and Fc? SOLUTION: ac = 4 2 r / T 2 = 4 2 (3) / (5)2 = 4.74 m s-2. Fc = mac = (2.5)(4.74) = 11.9 n. The tension is causing the centripetal force, so the tension is Fc = 11.9 n. Albert the Physics Cat © 2006 By Timothy K. Lund

Angular displacement and arc length Consider the rotating arm which has 6 paint cans along its radius. Each can has a spout that is opened for exactly a quarter of a revolution. We call  the angular displacement. All 6 color trails represent the same angular displacements of 90˚. Each color traces out a different displacement s. We call s the arc length. All 6 color trails represent different arc lengths.  s s s s s © 2006 By Timothy K. Lund

Angular displacement and arc length At this point it is useful to define a new way to measure angles – called radians. Looking at the above conversions we see that there are 2 rad in 360˚.  rad = 180° = 1/ 2 rev radian-degree-revolution conversions 2 rad = 360° = 1 rev EXAMPLE: Convert 30 into radians (rad) and convert 1.75 rad to degrees. SOLUTION:  30(  rad / 180° ) = 0.52 rad.  1.75 rad ( 180° /  rad ) = 100°. © 2006 By Timothy K. Lund

Angular displacement and arc length The relationship between angular displacement  and arc length s is where r is the radius.  rad = 180° = 1/ 2 rev radian-degree-revolution conversions 2 rad = 360° = 1 rev s = r  relation between s and   in radians EXAMPLE: Suppose the red line is located at a radius of 1.50 m and the green line is located at 1.25 m. Find their lengths. SOLUTION: 90(  rad / 180°) = 1.57 rad. s = r  = 1.501.57 = 2.4 m. s = r  = 1.251.57 = 2.0 m. © 2006 By Timothy K. Lund

Angular speed and speed The arc length s is simply the displacement we learned about in Topic 2, and is the s that is in s = ut + (1/2) at 2. Because speed is v = s / t, we see that v = s / t (definition of speed) = ( r  ) / t (substitution) = r (  / t ) (associative property) = r  (define    / t ) Thus We call  the angular speed. s = r  relation between s and   in radians © 2006 By Timothy K. Lund v = r  relation between v and   =  / t (rad s-1)

Angular speed and speed v = r  relation between v and   =  / t (rad s-1) EXAMPLE: Consider the following point mass moving at a constant speed v in a circle of radius r as shown. Find … the period T of the point mass, and (b) the frequency f of the point mass, and (c) the angular speed  of the point mass. SOLUTION: We need a time piece. For one revolution the period is T = 12 s. Frequency f = 1 / T = 1 / 12 = s. Angular speed is  =  / t = 2 rad / 12 s = 0.52 rad s-1. r v © 2006 By Timothy K. Lund

Angular speed and speed v = r  relation between v and   =  / t (rad s-1) EXAMPLE: Find the angular speed of the second hand on a clock. Then find the speed of the tip of the hand if it is 18.0 cm long. SOLUTION: A second hand turns 2 rad each 60 s. Thus it has an angular speed given by  = 2 / T = 2 / 60 = rad s-1. The speed of the tip is given by v = r  = 0.180(0.105) = ms-1. © 2006 By Timothy K. Lund FYI Speed depends on length or position but angular speed does not.

Angular speed and speed v = r  relation between v and   =  / t (rad s-1) EXAMPLE: A car rounds a ° turn in 6.0 seconds What is its angular speed during the turn? SOLUTION: Since  needs radians we begin by converting :  = 90°(  rad / 180° ) = 1.57 rad. Now we use  =  / t = 1.57 / 6.0 = 0.26 rad s-1. © 2006 By Timothy K. Lund

Banking The car is able to round the curve because of the friction between tire and pavement. The friction always points to the center of the circle. So, how does a plane follow a circular trajectory? There is no sideways friction force that the plane can use because there is no solid friction between the air and the plane. © 2006 By Timothy K. Lund

Banking Using control surfaces on the tail and the main wings, planes can execute three types of maneuver: ROLL – Ailerons act in opposing directions YAW – Tail rudder turns left or right PITCH – Ailerons and horizontal stabilizer act together © 2006 By Timothy K. Lund FYI It is the ROLL maneuver that gives a plane a centripetal force as we will see on the next slide.

Banking Even though cars use friction, roads are banked so that the need for friction is reduced. Instead of a component of the LIFT force providing a centripetal force, a component of the NORMAL force does so. © 2006 By Timothy K. Lund FYI A banked curve can be designed so that a car can make the turn even if it is perfectly frictionless! R FC W

Angular speed and centripetal acceleration Sometimes the angular speed of an object in circular motion is given, rather than its velocity. From v = r  we get v 2 = r 2  2. From ac = v 2 / r we get ac = r 2  2 / r ac = r  2. Putting it all together we have © 2006 By Timothy K. Lund ac = v 2 / r ac and Fc (all three forms) ac = 4 2 r / T 2 ac = r  2 Fc = mv 2 / r Fc = 4 2 mr / T 2 Fc = m  2 r

Angular velocity As speed with a direction is called velocity, angular speed with a direction is called angular velocity. To assign a direction to a rotation we use a right hand rule as follows: Rest the heel of your right hand on the rotating object. Make sure your fingers are curled in the direction of rotation. Your extended thumb points in the direction of the angular velocity.  = 2 / T = 2f =  / t relation between , T and f r v  © 2006 By Timothy K. Lund FYI Angular velocity always points perpendicular to the plane of motion!

Angular velocity  = 2 / T = 2f =  / t relation between , T and f PRACTICE: Find the angular velocity (in rad s-1) of the wheel on the shaft It is rotating at 30.0 rpm (revolutions per minute). SOLUTION: The magnitude of  is given by  = (30.0 rev / 60 s)(2 rad / rev) = 3.14 rad s-1. The direction of  is given by the right hand rule: “Place heel of right hand so fingers are curled in direction of rotation. Thumb gives the direction.” © 2006 By Timothy K. Lund

Identifying the forces providing centripetal forces PRACTICE: Identify at least five forces that are centripetal in nature: SOLUTION: The tension force (Albert the physics cat and Arnold). The friction force (the race car making the turn). The gravitational force (the baseball and the earth). The electric force (an electron orbiting a nucleus). The magnetic force (a moving charge in a B-field). © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems PRACTICE: Dobson is watching a 16-pound bowling ball being swung around at 50 m/s by Arnold. If the string is cut at the instant the ball is next to the ice cream, what will the ball do? (a) It will follow path A and strike Dobson's ice cream. (b) It will fly outward along curve path B. (c) It will fly tangent to the original circular path along C. © 2006 By Timothy K. Lund A C B

Solving centripetal acceleration and force problems EXAMPLE: Suppose a kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of RE = m, find the speed of the ball. SOLUTION: The ball is traveling in a circle of radius r = m. Fc is caused by the weight of the ball so that Fc = mg = (0.5)(10) = 5 n. Since Fc = mv 2 / r we have 5 = (0.5)v 2 / v = 8000 m s-1! © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems EXAMPLE: Suppose a kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). How long will it take the ball to return to Everest? SOLUTION: We want to find the period T. We know that v = 8000 m s-1. We also know that r = m. Since v = 2r / T we have T = 2r / v T = 2( )/ 8000 = (5030 s)(1 h / 3600 s) = 1.40 h. © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems EXAMPLE: Explain how an object can remain in orbit yet always be falling. SOLUTION: Throw the ball at progressively larger speeds. In all instances the force of gravity will draw the ball toward the center of the earth. When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface. The ball is effectively falling around the earth! © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems PRACTICE: Find the angular speed of the minute hand of a clock, and the rotation of the earth in one day. SOLUTION: The minute hand takes 1 hour to go around one time. Thus  = 2 / T = 2 / 3600 s = rad s-1. The earth takes 24 h for each revolution so that  = 2 / T = ( 2 / 24 h )( 1 h / 3600 s ) = rad s-1. This small angular speed is why we can’t really feel the earth as it spins. © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems EXAMPLE: The Foucault pendulum is a heavy pendulum on a very long cable that is set in oscillation over a round reference table. Explain how it can be used to tell time. SOLUTION: The blue arcs represent the motion of the pendulum bob relative to the universe at large. The the green lines represent the plane of motion of the pendulum relative to the building. © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems EXAMPLE: The Foucault pendulum is a heavy pendulum on a very long cable that is set in oscillation over a round reference table. Explain how it can be used to tell time. SOLUTION: Since the building is rotating with the earth at  = rad s-1, each hour the green line rotates by  = t = (3600) = rad (360/ 2 rad) = 15.0. © 2006 By Timothy K. Lund FYI This solution only works when the pendulum is at one of the poles. See the Wiki for a general solution.

Solving centripetal acceleration and force problems  90˚ EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Recall that  = rad s-1 anywhere on the earth. The blue arcs represent the lines of latitude. The white line R represents the earth’s radius. The yellow line r represents the radius of the circle a point at a latitude of  follows. Note that r = R cos , and that at the equator,  = 0˚ and at the pole,  = 90˚. r  R  0˚ © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems  EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Recall that  = rad s-1 anywhere on the earth. Thus, at the equator, r = R, and at the pole, r = 0. Furthermore, R = m. Then, at the equator, ac = r 2 =  = ms-2. Then, at the pole, ac = r 2 = 0 = ms-2. r  R  © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole. SOLUTION: Make a free-body diagram at the equator… Scales read the normal force R: F = ma R – W = - mac R = W – mac Then, R = 882 – ( 882 / 9.8 )  = 879 N. The man has apparently “lost” about 3 N! W R ac © 2006 By Timothy K. Lund

Solving centripetal acceleration and force problems © 2006 By Timothy K. Lund Use F = kx (k = CONST). kx = FC = mv 2/ r implies that as v increases, so does the centripetal force FC needed to move it in a circle. Thus, x increases.

Solving centripetal acceleration and force problems © 2006 By Timothy K. Lund kx = F  k = F / x = 18 / = 1800 Nm-1. FC = kx = 1800( – ) = 27 N. FC = v 2/ r  v 2 = r FC = 0.265(27) = 7.155 v = 2.7 ms-1.

Use v = r ( = CONST). Use a = r 2 ( = CONST). Solving centripetal acceleration and force problems © 2006 By Timothy K. Lund At P r = R v = R a = R  2 At Q r = 2R v = 2R = 2v a = 2R 2 = 2a

Topic 6: Circular motion and gravitation 6.1 – Circular motion

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Topic 6: Circular motion and gravitation 6.1 – Circular motion

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